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Marysya12 [62]
3 years ago
10

What is the change in gravitational potential energy of an apple with a mass of 0.25 kg if it falls 0.7 m from a table. Estimate

g to 9.81 ms2.
PLS HELP ASAP
Chemistry
1 answer:
Drupady [299]3 years ago
8 0

Answer:

1.71675 J

Explanation:

Please mark Brainliest?

Hope this helps!

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Ainat [17]
On the first one it is supposed to be 18. when you have a +1 charge you subtract it once. how i got 18 tho was from the protons. there was 19 so i subtracted that with 1 and got 18. hope that helped! :)

btw i’m not the best at explaining, i’m sorry :/
8 0
3 years ago
How much energy is required to vaporize 155 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol
netineya [11]

The energy required to vaporize 155 g of butane at its boiling point: 61,723 kJ

<h3>Further explanation</h3>

Enthalpy is the amount of system heat at constant pressure.

The enthalpy is symbolized by H, while the change in enthalpy is the difference between the final enthalpy and the initial enthalpy symbolized by ΔH.

\large{\boxed{\boxed{\bold{\Delta H=H_{End}-H_{First}}}}

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The standard unit is kilojoules (kJ)

The enthalpy change symbol (ΔH) is usually written behind the reaction equation.

Change in Standard Evaporation Enthalpy (ΔH vap) is a change in enthalpy at the evaporation of 1 mol liquid phase to the gas phase at its boiling point and standard pressure.

Examples of water evaporation:

 H₂O (l) ---> H₂O (g); ΔH vap = + 44kJ

The enthalpy of evaporation is positive because its energy is needed to break the attraction between molecules in a liquid

  • 155 g of butane

relative molecular mass of butane (C₄H₁₀) = 4.12 + 10.1 = 58 gram / mol

tex]\large{\boxed{mole\:=\:\frac{grams}{relative\:molecular\:mass}}}[/tex]

\large mole\:=\:\large \frac{155}{58}

mole = 2,672

Since the heat of vaporization for butane is 23.1 kj / mol, the energy needed to evaporate 2,672 moles of butane is:

23.1 kJ / mol x 2,672 mol = 61,723 kJ

<h3>Learn more</h3>

the heat of vaporization

brainly.com/question/11475740

The latent heat of vaporization

brainly.com/question/10555500

brainly.com/question/4176497

Keywords: the heat of vaporization, butane, mole, gram, exothermic, endothermic

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Bc it’s the only group that distinguishes amino acids from one another’s
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Will mark brainliest​
MArishka [77]

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D. water

<h2><em>H</em><em>ope it will help you</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em></h2>
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